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irakobra [83]
3 years ago
7

Tech A says that a gear set that has a drive gear with 9 teeth and a driven gear with 27 teeth has a gear ratio of 3:1. Tech B s

ays that the drive gear is also called the output gear. Who is correct?
Engineering
1 answer:
choli [55]3 years ago
5 0

Answer:

Tech A is correct.

Explanation:

Gears are toothed wheels that can be used to transmit power. When two or more gears are in tandem, a gear train is formed.

Gear ratio = \frac{number of teeth of the driven gear}{Number of teeth of the driving gear}

                = \frac{27}{9}

                = \frac{3}{1}

Gear ratio = 3:1

The driver gear is called the input gear since it transfers its power to the driven gear. While the driven gear is called the output gear because it produces an effect due to both gears.

Tech A is correct.

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What is true about a point and shoot camera?
mario62 [17]

Answer:

What is true about a point and shoot camera?

  • They are often smaller than SLR cameras

6 0
3 years ago
Find the total present worth of a series of cash flows with an annual interest rate of 2% per year. Round your answer to the nea
prisoha [69]

The total present worth is $19,783.01

The present worth of a series of cash flow is the value of the cash flows in year 0 (today)

Cash flow in year 0 = 5330

Cash flow in year 1 = 0

Cash flow in year 2 = 0

Cash flow in year 3 = 13075 / (1.02)^3 = 12,320.86

Cash flow in year 4 = 2308 / (1.02)^4 = 2,132.24

Present worth = $19,783.01

A similar question was solved here: brainly.com/question/9641711?referrer=searchResults

5 0
3 years ago
Two wastewater treatment plant workers (one male and one female) are exposed to hydrogen sulfide in confined spaces in the treat
Vlada [557]

Answer:

Go to explaination for the details of the answer.

Explanation:

In order to determine the lifetime (75 years) chronic daily exposure for each individual, we have to first state the terms of our equation:

CDI = Chronic Daily Intake

C= Chemical concentration

CR= Contact Rate

EFD= Exposure Frequency and Distribution

BW= Body Weight

AT = Average Time.

Having names our variables lets create the equations that will be used to derive our answers.

Please kindly check attachment for details of the answer.

5 0
3 years ago
A closed, rigid tank is lled with a gas modeled as an ideal gas, initially at 27°C and a gage pressure of 300 kPa. The gas is he
sergejj [24]

Answer:

T₂ =93.77  °C

Explanation:

Initial temperature ,T₁ =27°C= 273 +27 = 300 K

We know that

Absolute pressure = Gauge pressure + Atmospheric pressure

Initial pressure ,P₁ = 300+1=301 kPa

Final pressure  ,P₂= 367+1 = 368  kPa

Lets take  temperature=T₂

We know that ,If the volume of the gas is constant ,then we can say that

\dfrac{P_2}{P_1}=\dfrac{T_2}{T_1}

{T_2}=T_1\times \dfrac{P_2}{P_1}

Now by putting the values in the above equation we get

{T_2}=300\times \dfrac{368}{301}\ K

The temperature in  °C

T₂ = 366.77 - 273  °C

T₂ =93.77  °C

8 0
3 years ago
. Using the Newton Raphson method, determine the uniform flow depth in a trapezoidal channel with a bottom width of 3.0 m and si
Over [174]

Answer:

y  ≈ 2.5

Explanation:

Given data:

bottom width is 3 m

side slope is 1:2

discharge is 10 m^3/s

slope is 0.004

manning roughness coefficient is 0.015

manning equation is written as

v =1/n R^{2/3} s^{1/2}

where R is hydraulic radius

S = bed slope

Q = Av =A 1/n R^{2/3} s^{1/2}

A = 1/2 \times (B+B+4y) \times y =(B+2y) y

R =\frac{A}{P}

P is perimeter =  (B+2\sqrt{5} y)

R =\frac{(3+2y) y}{(3+2\sqrt{5} y)}

Q = (2+2y) y) \times 1/0.015 [\frac{(3+2y) y}{(3+2\sqrt{5} y)}]^{2/3} 0.004^{1/2}

solving for y100 =(2+2y) y) \times (1/0.015) [\frac{(3+2y) y}{(3+2\sqrt{5} y)}]^{2/3} \times 0.004^{1/2}

solving for y value by using iteration method ,we get

y  ≈ 2.5

5 0
3 years ago
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